It is surprising that in statistical physics as that of Fermi, Dirac, Bose and Einstein there is no use of statistical weight but in place the notion of entropy. Indeed a set of particles need a balance between the particles with a large energy and those with a small energy, like we need a lot of poor people to balance a rich one. There is therefore a maximum of statistical weight determining the probability of the values around average value.
Statistical weight and entropy are so closely related that one cannot exist without the other. Maximum entropy corresponds to the most unbiased distribution of statistical weights. If nothing is known except that a system must be in one of its possible states, and that the sum of the probabilities of it being in its possible states is unity, then the most unbiased distribution of statistical weights is a uniform one, with the entropy being k ln N if there are N possible states. If in addition to that it is also known that the system is in contact with a (very large) heat reservoir, then the most unbiased distribution of statistical weights is a Boltzmann distribution for the system alone, but still a uniform distribution for the combined (system plus heat reservoir).
During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
The so-called entropy was such a concept that was derived by mistake in history.
Calculus is not "take for granted".
In fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
It is well known that calculus has a definition.
Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.
Based on the definition of calculus, we know:
to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.
As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,
∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.
1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.
2) On the other hand, In fact, Q=f(P, V, T), then
∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).
We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.
that is, the so-called "entropy " doesn't exist at all.
Dear Jack Denur and Shufeng Zhang,
In the classical approach of statistical thermodynamics it was use the position and the momentum of the atoms of a gas. This is not useful since the alone variable is the heat. So fifty years ago I start a new determination. The last version taking into account the notion of perturbation is “Perturbations and Statistical distribution of the Thermal Energy”. In fact I never use the notion of Entropy and replace it supposing that the maximum of the statistical weight is obtain around the mean value U of the thermal energy that is 3/2kT for the gas. The distribution respect the balance between the high and the small values.
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